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24

You're actually hitting on a very famous concept here that revolutionized physics!! Your understanding is almost wholly correct and your analogy is a good one - excellent reasoning - the only thing missing is radiation from the system. This latter lack is mostly irrelevant for the level of question you have been thinking about: but I'll address that below. ...


20

Nerd Sniping! The answer is $\frac{4}{\pi} - \frac{1}{2}$. Simple explanation: http://www.mbeckler.org/resistor_grid/ Mathematical derivation: http://www.mathpages.com/home/kmath668/kmath668.htm


17

First, Field strength. This calculation is strictly an electric potential calculation; radiation and induction are safely ignored at 50Hz.* For a 200kV transmission line 20m above ground, the max electric field at ground level is about 1.2 kV/m. This number is reduced from the naive 200kV/20m=10 kV/m calculation by two effects: 1) The ~1/r variation in ...


14

It's not true. To see this, you can try an experiment with some batteries and light bulbs. Hook up two bulbs of different wattages (that is, with different resistances) in parallel with a single battery: ------------------------------------------ | | | Battery Bulb 1 Bulb 2 | ...


14

The physical 'meaning' of the imaginary part of the impedance is that it represents the energy storage part of the circuit element. To see this, let the sinusoidal current $i = I\cos(\omega t)$ be the current through a series RL circuit. The voltage across the combination is $$v = Ri + L\frac{di}{dt} = RI\cos(\omega t) - \omega LI\sin(\omega t)$$ The ...


13

where does that electricity go? The photons from the sun have energy and momentum, but not "electricity". Essentially, a photon (solar or otherwise) striking the solar panel can create an electron-hole pair (EHP) and, if the EHP is within or near the depletion zone, the pair will be separated by the built-in electric field. This results in a ...


12

Most probably yes; wireless devices are not grounded, so they are not lighting rods of any kind as it is frequently assumed. There are some theories that cell phones somehow attracts lightnings by the field they produce, but the theory behind is weak. Experimental evaluation is very hard, since lightning hits are quite rare, such events are guided by ...


12

Yes, it is possible. For example Kevin Brown did here and here including this table. so for the xkcd problem the answer is $-\frac{1}{2}+\frac{4}{\pi} \approx 0.773$.


12

$\def\vE{{\vec{E}}}$ $\def\vD{{\vec{D}}}$ $\def\vB{{\vec{B}}}$ $\def\vJ{{\vec{J}}}$ $\def\vr{{\vec{r}}}$ $\def\vA{{\vec{A}}}$ $\def\vH{{\vec{H}}}$ $\def\ddt{\frac{d}{dt}}$ $\def\rot{\operatorname{rot}}$ $\def\div{\operatorname{div}}$ $\def\grad{\operatorname{grad}}$ $\def\rmC{{\mathrm{C}}}$ $\def\rmM{{\mathrm{M}}}$ $\def\ph{{\varphi}}$ ...


11

AC or DC, you only get electrocuted if current passes through your body. (Current passing through any part of your body can be dangerous, and possibly cause an electrical burn, but current passing across your heart is the one that's really dangerous.) Touching just one wire at a time gives the current nowhere much to go. You are right to think that some ...


11

Alfred got in before me, but I have a diagram! I've marked all continuous bits of wire in the same colour, and marked the corresponding colours on the ends of the resistors. A quick redraw later and I get: which is a lot simpler!


10

Electric current, by definition, is a flow of charged particles. When someone says it is the propagation of the electric field, usually he means the following: The velocity of the electrons in the wires is very slow (few cm/s if I remember it right), but when one turn on the light he doesn't see any delay. The lamp starts lighting when the electrons start ...


10

Yes Sam, there definitely is electric field reshaping in the wire. Strangely, it is not talked about in hardly any physics texts, but there are surface charge accumulations along the wire which maintain the electric field in the direction of the wire. (Note: it is a surface charge distribution since any extra charge on a conductor will reside on the ...


10

They are in series circuit, so breaking one bulb breaks the circuit itself:


10

Batteries do not behave in such an ideal way across all conditions. The simplest model of a battery as a circuit element is the one you describe - a pure voltage source. A slightly-more sophisticated model is as a voltage source connected to a fixed resistor, called the battery's internal resistance. A typical battery has an internal resistance of between 1 ...


10

instead of thinking your body is empty and that a charged wire has to push electrons one by one through you and into the ground (blood is actually full of charge carriers), a better analogy would be a very long queue of pushy people. if the entrance to the apple store doesn't open, it doesn't matter how hard the guy at the back pushes--nothing moves. ...


10

If you redraw your diagram as: It should be clear which capacitors are in parallel and which are in series.


9

First, your camera is not designed to work with batteries below a certain voltage. When it detects an excessively low battery voltage it turns itself off. That circuit stays in the "off" state until voltage is completely removed from the circuit. When you operate your camera, the current required by your camera varies according to what you do with it. So ...


9

If the power line is 20m high, and has the voltage of 1MV , then the electric field (near ground), very roughly, is on order of 1000/30 kv ~ 30 000 v/m (the numbers are very approximate and the field is complicated because it is a wire near a plate scenario, and wire diameter is unknown but not too small else the air would break down, i.e. spark over, near ...


9

A human body may reflect and absorb radio frequencies, though not very efficiently. It may as well act as a resonance chamber for certain frequencies. For a signal of 100 MHz, the involved wavelength is 3 m, and so it is possible that parts of your body are acting slightly as a resonant chamber. (for an optimal resonance, you should have 1.5 m diameter, too ...


9

Why are wires in simple circuits approximated as equipotentials? Because one of the three assumptions of circuit theory is: All electrical effects happen instantaneously throughout a circuit. If the circuit is small enough compared to the wave length of the signals applied, all electric signals travel through it so quickly, that we can assume that they ...


9

The resistance of water, even with ions and minerals and such, is still fairly high. So, a tiny current flowed through the water, but not very much. Additionally, the heating effect that often destroys them when short circuited would also be nullified by the cooling water.


8

Although the question is not clear, my guess is that you are confused with the flow of current and mean position of electrons. In case of DC, we have a continuous flow of charge from one point to another point in the conductor, any electron completes a cycle of circuit. In case of AC, there is no net displacement of charge and this may lead one in ...


8

Electrons that reach the positive terminal indeed remain there. The potential difference between the two terminals pushes electrons from the negative anode toward the positive cathode. When an electron reaches the cathode, it stays there to equalize the original charge imbalance between the two nodes. When electrochemical redox reaction sustaining the ...


8

Sine and cosine waves are, physically, the most common. They are definitely the best description to what comes out of a wall socket, not because we like them mathematically, but because it's what comes out; electromotive force is generated in the power plant as a sinusoidal pattern with frequency 50/60 Hz. In the usual kind of generator, this is because in ...


8

Actually, induction works, although it is often used a bit differently than you described. You can place a warm superconductor loop into a normal coil. As you switch the coil on, there will be some current inside the superconductor, but since it is not cold yet, this current quickly dies down. Then you cool the superconductor below its critical temperature. ...


8

HINT: Notice that $$R_{eq}=R+\frac1{\frac1R+\frac1{R_{eq}}}$$


8

I find this sort of thing becomes much more intuitive if you can think of an analogy in terms of water. In this case, we can think of it like this: Here we have water flowing through a hole in a bath tub, into another tub underneath. The stick figure has been given the task of keeping the water level constant, by lifting water back up into the top tub ...


7

I'll give the answer to this question using an unusual method that showed up in the American Mathematical Monthly's problem section perhaps in the late 1970s. This is not necessarily the easy way to solve the problem, but it works out nicely from an algebraic point of view. The way most people solve most resistance problems is to use series and parallel ...


7

"Ground" refers to a particular voltage, generally taken to be "zero", or the voltage of the earth. A "virtual ground" is a wire in a circuit whose voltage is held to be zero not because it is directly connected to the true ground, but instead because it is actively driven to that voltage typically by feedback mechanisms. Here's an example of a virtual ...



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